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in two concentric circles, prove that all chords of the outer circle which touches the inner circle are of equal lengths.

 Hi Aris!

Here is the answer to your question.
 
Consider two concentric circles with centres at O. Let AB and CD be two chords of the outer circle which touch the inner circle at the points M and N respectively.
 
 
 
To prove the given question, it is sufficient to prove
AB = CD. For this join OM, ON, OB and OD.
 
Let the radius of outer and inner circles be R and r respectively.
 
AB touches the inner circle at M.
∴ AB is a tangent to the inner circle
∴ OM⊥AB
⇒ BM = ½ of AB
⇒ AB = 2BM
 
Similarly ON⊥CD, and CD = 2DN

 
Hence proved
 

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 i got the answer so sorry

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Let R and r be the radii of outer and inner circles with common centre as O. PQ and RS are two chords of the outer circle which are tangents to inner circle.
OM and ON are the radii of inner circle, so these will be perpendicular to PQ and RS respectively. [Radius is perpendicular to the tangent at the point of contact]
 
In right-angled triangle MPO,
(PO)2 = (MP)2 + (MO)2 [By Pythagoras Theorem]
(R)2 = (MP)2 + (r)2
MP2 = R2 – r2
Thus, PQ = RS
 
Similarly all chords which are tangents to the inner circle are equal in length.
 
CHEERS
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 GOOD PROOF

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